As you have used #f(x)# I am assuming you are using Calculus.

#color(blue)("Short cut method")#

By sight: #"rate of change "-> (dy)/(dx)=2x#

At #x=0" "(dy)/(dx)=2(0)=0#

At #x=2" "(dy)/(dx)=2(2)=4#

Thus the average rate of change is

#(0+4)/2=2#

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#color(blue)("From first principles")#

Let #" "y=x^2+2#..........................(1)

Increment #x" by the very small amount of "deltax#

As #x# has change then #y# will change as well

Let the change in #y" be "deltay#

Now we have

#y+deltay=(x+deltax)^2+2#

#y+deltay=x^2+2xdeltax+2#.....................(2)

Subtract equation (1) from equation (2)

#y+deltay=x^2+2xdeltax+2#

#underline(y" "=x^2" "+2)# apply the subtraction

#" "deltay = 0""+2xdeltax+0#

Divide by #deltax#

#(deltay)/(deltax)=2x xx(deltax)/(deltax)#

but #(deltax)/(deltax)=1#

So

#lim_(deltaxto0) (deltay)/(deltax)= (dx)/(dy)=2x #

Then solve for average rate of change as above.

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Note that# lim_(deltaxto0)# means consider the situation where

#deltax# gets so small it may as well be zero, but in reality it has not quite got there!